timstirling:
I've been trying to find out what sort of motors would be required by my robot in order to carry around a laptop and some cameras. The tutoiral was helpful in revising my physics knowledge but I have a question about the units used.

Specifically with the Robot Motor Factor. The equation examples have the Torque in lb/in but the acceleration and velocity are in ft/s, i.e there is a mix of inches and feet going on. Is this correct, doesn't seem right to me unless I'm missing something.

Now, using similar velocities as in the example equations= 3ft/s and 2ft/s^2~= 90cm/s and 60cm/s.Metric: 6*90*60/(2Pi)= 4751.... with your imperial equatgion and ft/s: 11.6.......

Why is the velocity in feet and the torque in inches?

Admin:
Oops! I wrote inches where I should have written feet in several places. I even had another mechanical engineer check before I posted this and she didnt even notice it . . . I guess I was inspired by NASA ;)

Anyway it should be corrected now.

I probably should have written this using the metric system . . . its just Im american and although it is easier I believe to calculate in metric, it is easier for me to think in the english system.

Those motors are way below your spec . . . you might just want a velocity of 1 ft/s and accel of 1 ft/s for a minimum RMF of 303 lb in rps. Then find a motor about 3x better RMF than motor 2. The problem you will find is that stronger motors are expensive and weight on your robot means money out of your pocket . . .

Anyway let me know if I pulled a NASA again . . . g'luck!

timstirling:
Ok, thanks. So I haven't gone bad.

This is one time when i wish i was wrong though. Looks like it is going to be very difficult to get hold of motors powerful enough to lug the laptop around. And if I do get motors powerful enough then the motor controllers I was hoping to use wont be able to take the current.